Question

$\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We have $cotA=\frac{cosA}{sinA}=\frac{2co{s}^{2}A}{2sinAcosA}=\frac{1+cos2A}{sin2A}$
Putting $A=7\frac{1^{\circ }}{2}\Rightarrow cot7\frac{1^{\circ }}{2}=\frac{1+cos{15}^{\circ }}{sin{15}^{\circ }}$
$=\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}}$
$=\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\frac{2\sqrt{6}+3+\sqrt{3}+2\sqrt{2}+\sqrt{3}+1}{3-1}$
$=\frac{2\sqrt{2}+2\sqrt{3}+4+2\sqrt{6}}{2}$
$=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

Hence, We get $\cot 7\frac{1^{\circ }}{2}=\sqrt{6}+\sqrt{2}+\sqrt{3}+\sqrt{4}.$